Question

A child weighs 160 N on the surface of the earth. Calculate the weight of the child at
height of 3.6*106 m from the surface of the earth, where radius of the earth is 6400
km.

Answer 1

Answer:

please mark as brainliest

Explanation:

see the photo

Answer 2

Explanation:

g = Acceleration due to Gravity = 9.8 m/sec^2

g = Acceleration due to Gravity = 9.8 m/sec^2g'= Acceleration due to Gravity at height (h)=?

g = Acceleration due to Gravity = 9.8 m/sec^2g'= Acceleration due to Gravity at height (h)=?h = height = 3.6 × 10^6m = 3.36 × 10^3 km

g = Acceleration due to Gravity = 9.8 m/sec^2g'= Acceleration due to Gravity at height (h)=?h = height = 3.6 × 10^6m = 3.36 × 10^3 kmR = Radius of earth = 6400km =6.4 × 10^3 km

w= weight of child on earth surface =160 N

w'= weight of child at above earth surface = ?

w'= weight of child at above earth surface = ?g' = g[1- (2h/R)]

w'= weight of child at above earth surface = ?g' = g[1- (2h/R)]g= 9.8[1-(2×3.6×10^3÷6.4×10^3)]

w'= weight of child at above earth surface = ?g' = g[1- (2h/R)]g= 9.8[1-(2×3.6×10^3÷6.4×10^3)]g= |-1.225| m/sec^2

w'= weight of child at above earth surface = ?g' = g[1- (2h/R)]g= 9.8[1-(2×3.6×10^3÷6.4×10^3)]g= |-1.225| m/sec^2 = 1.225 m/sec^2

w'= weight of child at above earth surface = ?g' = g[1- (2h/R)]g= 9.8[1-(2×3.6×10^3÷6.4×10^3)]g= |-1.225| m/sec^2 = 1.225 m/sec^2w = m×g

w'= weight of child at above earth surface = ?g' = g[1- (2h/R)]g= 9.8[1-(2×3.6×10^3÷6.4×10^3)]g= |-1.225| m/sec^2 = 1.225 m/sec^2w = m×gw/w' = mg/mg' = g/g'

w'= weight of child at above earth surface = ?g' = g[1- (2h/R)]g= 9.8[1-(2×3.6×10^3÷6.4×10^3)]g= |-1.225| m/sec^2 = 1.225 m/sec^2w = m×gw/w' = mg/mg' = g/g'(w×g')÷g = w'

w'= weight of child at above earth surface = ?g' = g[1- (2h/R)]g= 9.8[1-(2×3.6×10^3÷6.4×10^3)]g= |-1.225| m/sec^2 = 1.225 m/sec^2w = m×gw/w' = mg/mg' = g/g'(w×g')÷g = w'w' = (160×1.225)÷(9.8)

w'= weight of child at above earth surface = ?g' = g[1- (2h/R)]g= 9.8[1-(2×3.6×10^3÷6.4×10^3)]g= |-1.225| m/sec^2 = 1.225 m/sec^2w = m×gw/w' = mg/mg' = g/g'(w×g')÷g = w'w' = (160×1.225)÷(9.8)w'= 20N

w'= weight of child at above earth surface = ?g' = g[1- (2h/R)]g= 9.8[1-(2×3.6×10^3÷6.4×10^3)]g= |-1.225| m/sec^2 = 1.225 m/sec^2w = m×gw/w' = mg/mg' = g/g'(w×g')÷g = w'w' = (160×1.225)÷(9.8)w'= 20NTherefore weight of child at 3.6 × 10^6 m will be 20N

Hope it helps mark me as brainliest